Velocity of the ring wraiths relative to the air:
[tex]\vec v_{W/A}=(v_{W/A}\cos45^\circ,v_{W/A}\sin45^\circ)=\left(\dfrac{v_{W/A}}{\sqrt2},\dfrac{v_{W/A}}{\sqrt2}\right)[/tex]
The speed [tex]v_{W/A}[/tex] is what we want to find - this is the speed at which the wraiths are flying in the air.
(Note: [tex]\vec v[/tex] denotes velocity, while [tex]v[/tex] denotes speed)
Velocity of air relative to the ground:
[tex]\vec v_{A/G}=((69\,\mathrm{mph})\cos180^\circ,(69\,\mathrm{mph})\sin180^\circ)=(-69,0)\,\mathrm{mph}[/tex]
The ring wraiths want their trajectory to be due north, which means their velocity relative to the ground should be
[tex]\vec v_{W/G}=(v_{W/G}\cos90^\circ,v_{W/G}\sin90^\circ)=(0,v_{W/G})[/tex]
To an observer on the ground, [tex]v_{W/G}[/tex] is the speed at which the wraiths would appear to be moving in the air.
The relative velocities satisify
[tex]\vec v_{W/A}+\vec v_{A/G}=\vec v_{W/G}[/tex]
[tex]\implies\begin{cases}\dfrac{v_{W/A}}{\sqrt2}-(69\,\mathrm{mph})=0\\\\\dfrac{v_{W/A}}{\sqrt2}=v_{W/G}\end{cases}[/tex]
[tex]\implies v_{W/A}=69\sqrt2\,\mathrm{mph}\approx98\,\mathrm{mph}[/tex]
